《The linear algebra a beginning graduate student ought to know》 Page 8 Field $a+b\sqrt{p}$($p$ is a prime)

The set $\bf{Q}^2$ is a subfield of the field $\bf{C}$ defined above.However, it is also possible to define field structures on $\bf{Q}^2$ in other ways.Indeed, let $F = \bf{Q}^2$ and let $p$ be a fixed prime integer. Define addition and multiplication on $F$ by setting $(a, b) + (c, d) = (a + c, b + d)$ and $(a, b)\cdot(c, d) = (ac + bdp, ad + bc)$.

Now I prove this forms a field.

Proof:

1.$\bf{Q}^2$ together with + form an abelian group.

2.$(0,0)$ is the additive zero in $\bf{Q}^2$.

3.Now I prove that $\bf{Q}^2\backslash\{(0,0)\}$ is an abelian group under multiplication.Now we calculate its multiply identity.
\begin{equation}
\begin{cases}
ac+bdp=a\\
ad+bc=b\\
\end{cases}
\end{equation}

So
\begin{equation}
c=\frac{\begin{vmatrix}
a&bp\\
b&a\\
\end{vmatrix}}{\begin{vmatrix}
a&bp\\
b&a\\
\end{vmatrix}}=1
\end{equation}
\begin{equation}
d=\frac{\begin{vmatrix}
a&a\\
b&b\\
\end{vmatrix}}{\begin{vmatrix}
a&bp\\
b&a\\
\end{vmatrix}}=0
\end{equation}
So if it has a multiply identity,this identity would be $(1,0)$.Now I examine that $(1,0)$ is actually a multiply identity of $\bf{Q}^2\backslash\{0\}$.This is very easy to verify.

Now I would like to find out multiply inverse for every element of $\bf{Q}\backslash\{0\}$.

\begin{equation}
(a,b)\cdot(m,n)=(1,0)
\end{equation}

So
\begin{equation}
m=\frac{\begin{vmatrix}
1&bp\\
0&a\\
\end{vmatrix}}{\begin{vmatrix}
a&bp\\
b&a\\
\end{vmatrix}}=\frac{a}{a^2-b^2p}
\end{equation}
\begin{equation}
n=\frac{\begin{vmatrix}
a&1\\
b&0\\
\end{vmatrix}}{\begin{vmatrix}
a&bp\\
b&a\\
\end{vmatrix}}=\frac{-b}{a^2-b^2p}
\end{equation}
(I need to point out that $a^2-b^2p\neq 0$,this is because if $a^2=b^2p$,then $p$ is not a prime,which is a contradiction.)Now I examine $(\frac{a}{a^2-b^2p},\frac{-b}{a^2-b^2p})$ is actually a multiply inverse of $(a,b)$.I just need to examine that
\begin{equation}
(\frac{a}{a^2-b^2p},\frac{-b}{a^2-b^2p})\cdot (a,b)=(1,0)
\end{equation}
This is easy to verify.

Now I prove the commutivity:This is easy to verify.

Now I prove the associativity:
\begin{equation}
[(a,b)\cdot(c,d)]\cdot(e,f)=(a,b)\cdot[(c,d)\cdot (e,f)]
\end{equation}
\begin{equation}
[(a,b)\cdot(c,d)]\cdot(e,f)=(ace+bdep+adfp+bcfp,ade+bce+acf+bdfp)
\end{equation}
Let $a\leftrightarrow e$,$b\leftrightarrow f$,then $(ace+bdep+adfp+bcfp,ade+bce+acf+bdfp)$ becomes itself.So the associativity holds.

So $\bf{Q}^2\backslash\{0\}$ form a abelian group with regard to multiplication.

So $\bf{Q}^2$ form a field.